 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-2/12f/solution

By the partition and the law of total probability,

P (A) = \sum_{j} P (A \cap B_{j}) = \sum_{j} P (A ∣ B_{j}) P (B_{j})

. Also

P (B_{i} ∣ A) = P (A \cap B_{i}) / P (A)

. Combining these identities proves Bayes theorem:

P (B_{i} ∣ A) = \frac{P ( A ∣ B _{i} ) P ( B _{i} )}{\sum _{j} P ( A ∣ B _{j} ) P ( B _{j} )} .

The assumed positive probabilities make the conditional expressions well defined.

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